A) 255
B) 127
C) 60
D) None of these
Correct Answer: A
Solution :
Given, \[(1+x)(1+{{x}^{2}})(1+{{x}^{4}})\]???.\[(1+{{x}^{128}})\] \[=1+x+{{x}^{2}}+....+{{x}^{n}}\] \[\therefore \]\[(1-x)\{(1+x)(1+{{x}^{2}})......(1+{{x}^{128}})\}\] \[=1-{{x}^{n+1}}\] \[\Rightarrow \]\[(1-{{x}^{2}})\{1+{{x}^{2}}......(1+{{x}^{128}})\}\]\[=1-{{x}^{n+1}}\] \[\Rightarrow \]\[(1-{{x}^{4}}).....(1+{{x}^{128}})=1-{{x}^{n+1}}\] \[\Rightarrow \]\[1-{{x}^{256}}=1-{{x}^{n+1}}\] \[\therefore \] \[n+1=256\] \[\Rightarrow \]\[n=255\]You need to login to perform this action.
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